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Sagot :

Jiya871

Given:-

  • HM = 2x+5
  • FM = X+8
  • HF = 3x+2
  • perimeter (∆HFM) = 51 units

To find:-

  • whether ∆ HFM is congruent to ∆FMK

Solution:-

:[tex]\implies[/tex] perimeter (∆HFM) = HM+FM+HF

:[tex]\implies[/tex] 51 = 2x+5+x+8+3x+2

:[tex]\implies[/tex] 51 = 6x+15

:[tex]\implies[/tex] 36 = 6x

:[tex]\implies[/tex] x = 6 units.

in ∆HFM,

  • HM = 2x+5 = 2(6)+5 = 12+5 = 17units
  • FM = X+8 = 6+8 = 14 units
  • HF = 3x+2 = 3(6)+2 = 20 units

in ∆KFM,

  • FM = 14 units
  • KF = 4x-3 = 4(6)-3 = 21 units
  • MK = 3x-1 = 3(6)-1 = 17 units

Since, all the sides of one triangle does not correspond to the other, these traingles are not congruent.

Hope it helps ⭐⭐⭐

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